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Mathematics

Geometry & Trigonometry

Welcome to Geometry & Trigonometry — the module where numbers meet shapes. Whether it is finding the distance between two points, writing the equation of a circle, or solving a triangle with the Law of Cosines, this module gives you the tools to work with space and angles.

You will start with coordinate geometry (points, lines, distances), progress through circles, the unit circle, trigonometric identities and the Laws of Sines and Cosines, then the conics (parabolas, ellipses and hyperbolas) and vectors, and finish with complex numbers and solid geometry. The syllabus lists analytic geometry, vectors, complex numbers and solid geometry under Geometry and Algebra, and trigonometric functions under Functions; these topics also connect naturally to physics and engineering.

Grab your protractor (just kidding — you will not need one!) and let's begin.

What the Exam Tests

The syllabus calls this area Geometry and Algebra: plane analytic geometry (the equations and properties of lines, circles, ellipses, hyperbolas and parabolas), vectors and complex numbers, and solid geometry (coordinates in space and the properties of simple solids). This module also teaches trigonometric identities and solving triangles with the laws of sines and cosines.

Questions can ask for the distance from a point to a line, the centre and radius of a circle, the eccentricity or foci of a conic, a dot product or the angle between two vectors, the modulus of a complex number, or a volume or surface area.

In our mock papers Geometry and Algebra gets 16 of the 48 questions. The CSCA publishes the topics but not how many questions each gets, so this split is our own reading of the syllabus.

Coordinate geometry (analytic geometry) places geometric shapes on the xy-plane so we can study them with algebra.

Key Formulas

  • Distance: d = √((x₂-x₁)² + (y₂-y₁)²). This is just the Pythagorean theorem applied to the horizontal and vertical differences.
  • Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2). Average the coordinates.
  • Gradient (Slope): m = (y₂-y₁)/(x₂-x₁). Rise over run.

Chinese textbooks call the gradient 斜率 and write it k, with k = tan α, where α is the 倾斜角 (inclination angle, 0° ≤ α < 180°). A vertical line (α = 90°) has no slope. Lines are written y = kx + b (b = 截距, the y-intercept).

Gradient Relationships

  • Parallel lines: same gradient → m₁ = m₂.
  • Perpendicular lines: gradients are negative reciprocals → m₁ · m₂ = -1.

Collinear Points
Three points are collinear (on the same line) if the gradient between any two pairs is equal.

Section Formula
The point dividing the segment from (x₁,y₁) to (x₂,y₂) in the ratio m:n is: ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).

These basics underpin every geometry question. The distance formula alone can determine whether a triangle is isosceles, right-angled, or equilateral.

Locus Problems
A locus is the set of all points that satisfy a stated geometric condition, and most locus questions are really conic questions in disguise. Learn the four standard conditions and the answer follows immediately:

  • Equidistant from two fixed points → the perpendicular bisector of the segment joining them. Points equidistant from (−2, 0) and (4, 0) form the vertical line x = 1.
  • A fixed distance from one fixed point → a circle.
  • Equidistant from a point and a line (the point not on the line) → a parabola, with the point as focus and the line as directrix.
  • The sum of the distances from two fixed points is constant and greater than the distance between them → an ellipse with those points as foci. The absolute difference being a non-zero constant less than the distance between them gives a hyperbola instead.

Check the condition. If the sum equals the distance between the points, the locus is only the segment joining them: |PF₁| + |PF₂| = 6 with F₁(−3, 0), F₂(3, 0) is the segment from (−3, 0) to (3, 0), not an ellipse. If the point lies on the line, "equidistant from point and line" gives the perpendicular line through the point, not a parabola.

So a point whose distances from (3, 0) and (−3, 0) always add to 10 traces an ellipse: 2a = 10 gives a = 5, and c = 3 gives b² = a² − c² = 16, so the locus is x²/25 + y²/16 = 1. When no standard condition fits, fall back on the method itself: write the condition as an equation in the distances from a general point (x, y), square to clear the roots, and simplify.

The Centroid of a Triangle
The three medians of a triangle — each joining a vertex to the midpoint of the opposite side — always meet at one point, the centroid, and in coordinates it is simply the average of the three vertices:

G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)

For vertices (0, 0), (6, 0) and (3, 9) the centroid is ((0+6+3)/3, (0+0+9)/3) = (3, 3). The centroid is the triangle’s balance point, and it divides each median in the ratio 2 : 1 from the vertex. The same averaging works with position vectors, so the centroid of A, B and C has position vector (a + b + c)/3.

Key takeaway: Distance = √((Δx)²+(Δy)²). Midpoint: average coordinates. Gradient: rise/run. Parallel: equal gradients. Perpendicular: m₁·m₂ = -1.

Key Formulas

d = √((x₂-x₁)²+(y₂-y₁)²)
M = ((x₁+x₂)/2, (y₁+y₂)/2)
m = (y₂-y₁)/(x₂-x₁)

Worked Example 1

Example 1: Find the distance between A(1, 3) and B(4, 7).

d = √((4-1)²+(7-3)²) = √(9+16) = √25 = 5.

Worked Example 2

Example 2: Find the midpoint of (-2, 6) and (4, -2).

M = ((-2+4)/2, (6+(-2))/2) = (1, 2).

Worked Example 3

Example 3: Line L₁ has gradient 2. Find the gradient of a line perpendicular to L₁.

m₂ = -1/m₁ = -1/2.

Exam tip: Label your points clearly as (x₁,y₁) and (x₂,y₂) before substituting into formulas.

Exam tip: To decide whether a triangle is right-angled, compute all three side lengths and check whether a² + b² = c².

Common mistake: Subtracting coordinates in wrong order: Be consistent — (x₂-x₁) and (y₂-y₁) must use the same order.

Common mistake: Forgetting the square root: d² = (Δx)²+(Δy)² — do not forget to take the root for the actual distance.

Try This Yourself

Find the distance between (0, 0) and (5, 12).

Practise Lines & Distances

Module Summary

In this module you explored coordinate geometry (distances, lines, circles), mastered trigonometry (unit circle, identities, Laws of Sines and Cosines), studied conic sections (parabolas, ellipses and hyperbolas), learned vector operations, and worked with complex numbers and solid geometry in three dimensions. These tools let you analyse shapes, solve triangles, and describe curves — skills the syllabus lists under Geometry and Algebra and that you will reuse in physics, engineering, and computer science.

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